[Numpy-discussion] numpy.nansum() behavior in 1.3.0

[josef.pktd at gmail.com]

On Mon, Jun 1, 2009 at 7:58 PM, Keith Goodman <kwgoodman at gmail.com> wrote:
> On Mon, Jun 1, 2009 at 4:50 PM,  <josef.pktd at gmail.com> wrote:
>> On Mon, Jun 1, 2009 at 7:43 PM,  <josef.pktd at gmail.com> wrote:
>>> On Mon, Jun 1, 2009 at 7:30 PM, Robert Kern <robert.kern at gmail.com> wrote:
>>>> On Mon, Jun 1, 2009 at 15:31,  <josef.pktd at gmail.com> wrote:
>>>>> On Mon, Jun 1, 2009 at 4:06 PM, Alan G Isaac <aisaac at american.edu> wrote:
>>>>>> On 6/1/2009 3:38 PM josef.pktd at gmail.com apparently wrote:
>>>>>>> Here's a good one:
>>>>>>>
>>>>>>>>>> np.isnan([]).all()
>>>>>>> True
>>>>>>>>>> np.isnan([]).any()
>>>>>>> False
>>>>>>
>>>>>>
>>>>>>  >>> all([])
>>>>>> True
>>>>>>  >>> any([])
>>>>>> False
>>>>>
>>>>> also:
>>>>>
>>>>>>>> y
>>>>> array([], dtype=float64)
>>>>>>>> (y>0).all()
>>>>> True
>>>>>>>> (y>0).any()
>>>>> False
>>>>>>>> ((y>0)>0).sum()
>>>>> 0
>>>>>
>>>>> I don't know what's the logic, but it causes the bug in np.nansum.
>>>>
>>>> You will have to special-case empty arrays, then.
>>>>
>>>
>>> is np.size the right check for non-empty array, including subtypes?
>>>
>>> i.e.
>>>
>>> if y.size and mask.all():
>>>        return np.nan
>>>
>>> or more explicit
>>> if y.size > 0 and mask.all():
>>>        return np.nan
>>>
>>
>> Actually, now I think this is the wrong behavior, nansum should never
>> return nan.
>>
>>>>> np.nansum([np.nan, np.nan])
>> 1.#QNAN
>>
>> shouldn't this be zero
>
> The doc string says it is zero: "Return the sum of array elements over
> a given axis treating Not a Numbers (NaNs) as zero." Treating NaNs
> differently in different cases is harder to explain.


http://projects.scipy.org/numpy/ticket/1123

open for review

Josef